Final answer:
An odd integer can be expressed in the form 2k + 1. Therefore, the square of any odd integer will have an even component when squared. Adding two such even components results in an even integer, proving that a² + b² is even for odd integers a and b.
Step-by-step explanation:
To prove that if a and b are any odd integers, then a² + b² is even, we need to express odd integers in a general form. Any odd integer can be written as 2k + 1, where k is any integer. So if a is an odd integer, it can be expressed as 2m + 1, and b as 2n + 1 where m and n are integers.
The square of an odd integer a would thus be (2m + 1)² = 4m² + 4m + 1, and similarly, for b, (2n + 1)² = 4n² + 4n + 1. When we add these two squares together, we get:
a² + b² = (4m² + 4m + 1) + (4n² + 4n + 1)
= 4m² + 4n² + 4m + 4n + 2
= 2(2m² + 2n² + 2m + 2n + 1)
Since 2m² + 2n² + 2m + 2n + 1 is an integer (as it is a sum and product of integers), multiplying it by 2 results in an even integer. Therefore, a² + b² is even, which proves our statement.